Suppose we have a vector field $f(x, y) = (\sqrt{x}, 2y)$ and a curve $C$ that is parameterized by $\alpha(t) = \left( t^2, 3t \right)$ for $0 < t < 2$. What is the line integral of $f$ along $C$ ? $ \int_C f \cdot d\alpha = $
Explanation: Given a vector field $f$, a parameterization $\alpha$, and bounds $t_0$ and $t_1$, we can calculate the line integral as follows: $ \int_C f \cdot d\alpha = \int_{t_1}^{t_2} f(\alpha(t)) \cdot \alpha'(t) \, dt$ Here, $f(x, y) = (\sqrt{x}, 2y)$ and $\alpha(t) = (t^2, 3t)$. $\begin{aligned} &f(\alpha(t)) = \left( t, 6t \right) \\ \\ &\alpha'(t) = (2t, 3) \end{aligned}$ Now we can rewrite our line integral as a single-variable integral. $ \int_C f \cdot d\alpha = \int_0^2 \left( t, 6t \right) \cdot (2t, 3) \, dt$ Let's solve the integral. $\begin{aligned} &\int_0^2 \left( t, 6t \right) \cdot (2t, 3) \, dt \\ \\ &= \int_0^2 2t^2 + 18t \, dt \\ \\ &= \left[ \dfrac{2}{3} t^3 + 9t^2 \right]_0^2 \\ \\ &= \dfrac{16}{3} + 36 \\ \\ &= \dfrac{124}{3} \end{aligned}$ In conclusion, the line integral $ \int_C f \cdot d\alpha = \dfrac{124}{3}$.